[R] linear model with similar response predictor

[David Winsemius]

On Aug 5, 2010, at 6:50 AM, Giuseppe Amatulli wrote:

> Hi,
> can somebody tell me why R is not able to calculate a linear model
> written in this way?
>
>> lm (seq(1:100)~seq(1:100))
>
> Call:
> lm(formula = seq(1:100) ~ seq(1:100))
>
> Coefficients:
> (Intercept)
>       50.5
>
> Warning messages:
> 1: In model.matrix.default(mt, mf, contrasts) :
>  the response appeared on the right-hand side and was dropped
> 2: In model.matrix.default(mt, mf, contrasts) :
>  problem with term 1 in model.matrix: no columns are assigned
>>
>
> Seems that is not able to dealing with similar number and so with very
> small numbers for calculating the coefficients. Moreover also
> Intercepts is wrong it should be equal to 0.

The group mean is correctly calculated.

>
> I compile the R 2.11.1 in the Fedora 13.
> Should i download a library?

That would not appear to be necessary:

 > x= 1:100
 > y=1:100
 > lm(y~x)

Call:
lm(formula = y ~ x)

Coefficients:
(Intercept)            x
  -5.684e-14    1.000e+00

> Some configure options are missing during the installation? or is a  
> bug?

Not sure. To determine that you would need to read the help page very  
carefully and determine whether putting such raw expressions into a  
location where formulas were generally used actually is allowed. A bit  
of experimentation shows that it is probably the rhs logic (as was  
also suggested by the warning message) where the issue arises.

 > lm (seq(1:100)~., data=list(a=1:100))

Call:
lm(formula = seq(1:100) ~ ., data = list(a = 1:100))

Coefficients:
(Intercept)            a
  -5.684e-14    1.000e+00

Read the help page for formula and note the section regarding the I()  
function:

 > lm(seq(1:100)~I(seq(1:100)) )

Call:
lm(formula = seq(1:100) ~ I(seq(1:100)))

Coefficients:
   (Intercept)  I(seq(1:100))
    -5.684e-14      1.000e+00



> Thanks in advance
> Giuseppe Amatulli

David Winsemius, MD
West Hartford, CT